Inequalities-5

Applying AM-GM inequality, we have:

$\dfrac{ab+bc+ca}{abc}=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{3}{3\sqrt[3]{abc}}\ge\dfrac{9}{a+b+c}=9$ .

The equality occurs only when $a=b=c=\dfrac{1}{3}$ .

So the answer is $\boxed{9}$ .

Let $f(x)=\frac{1}{x}$ . Since $f(x)$ is a convex function, by Jensen's Inequality, we have $f(\frac{a+b+c}{3}) \leq \frac{f(a)+f(b)+f(c)}{3}$ this simplifies to $\frac{1}{a}+\frac{1}{b}+\frac{1}{b} \geq 9$ . And hence, maximum value of the expression is $\boxed{9}$

Since a+b+c = 1, a=b=c=1/3, Then ab+bc+ca =1/a+1/b+1/c =(ab+bc+ca)/abc greater than (3/9)/1/27 = (3x27)/9. So the answer is greater than 9